# Problem 14.3b3

A First Course in String Theory

14.3 Massive level in the open superstring.

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What is a zero mode?

p.315 “Being zero modes, these creation operators do not contribute to the mass-squared of the states.”

How come there are creation operators that do not contribute to the mass-squared of the states?

Before each creation operator, there is a multiple which is the same as the absolute value of the index of the creation operators.

If an index can be zero, the corresponding term can be zero.

— Me@2015.09.26 08:44 PM

Consider the NS-sector:

Equation (14.37):

$M^2 = \frac{1}{\alpha'} \left( \frac{-1}{2} + N^\perp \right)$

$N^\perp = \sum_{p=1}^\infty \alpha_{-p}^I \alpha_p^I + \sum_{r=\frac{1}{2}, \frac{3}{2} ...} r b_{-r}^I b_r^I$

For the NS sector, the $r$ values in $b^I_{r}$ are half-integers, thus cannot be zero. So every creation operator $b_{-r}^I$ contributes to the mass-squared.

(p.312 “… the negatively moded coefficients $b_{-1/2}^I$, $b_{-3/2}^I$, $b_{-5/2}^I$, …, are creation operators, …”)

Consider the R-sector:

Equation (14.53):

$M^2 = \frac{1}{\alpha'} \sum_{n \ge 1} \left( \alpha_{-n}^I \alpha_{n}^I + n d_{-n}^I d_n^I \right)$

For the R sector, the $n$ values in $d^I_n$ are integers, thus can be zero. So some of the creation operators $d^I_{-n}$ are zero modes.

p.315 “… the eight fermionic zero modes $d_0^I$…”

— Me@2015-10-11 11:01:44 AM